Evaluate and Simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$












1














I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.



I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$



My Process:



$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$



$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$



$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$



$frac{6}{10} + frac{5}{10}$



$frac{11}{10}$






algebra-precalculus trigonometry







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  • 1




    Your $frac{11}{10}$ can't possibly be the cosine of anything!
    – TonyK
    Nov 22 '18 at 18:05






  • 1




    @TonyK Not for real ;^)
    – Michael Hoppe
    Nov 22 '18 at 19:28
















1














I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.



I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$



My Process:



$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$



$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$



$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$



$frac{6}{10} + frac{5}{10}$



$frac{11}{10}$






algebra-precalculus trigonometry







share|cite|improve this question


















  • 1




    Your $frac{11}{10}$ can't possibly be the cosine of anything!
    – TonyK
    Nov 22 '18 at 18:05






  • 1




    @TonyK Not for real ;^)
    – Michael Hoppe
    Nov 22 '18 at 19:28














1












1








1







I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.



I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$



My Process:



$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$



$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$



$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$



$frac{6}{10} + frac{5}{10}$



$frac{11}{10}$






algebra-precalculus trigonometry







share|cite|improve this question













I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.



I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$



My Process:



$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$



$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$



$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$



$frac{6}{10} + frac{5}{10}$



$frac{11}{10}$






algebra-precalculus trigonometry




algebra-precalculus trigonometry






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share|cite|improve this question











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asked Nov 22 '18 at 17:56









LuminousNutriaLuminousNutria

1709




1709








  • 1




    Your $frac{11}{10}$ can't possibly be the cosine of anything!
    – TonyK
    Nov 22 '18 at 18:05






  • 1




    @TonyK Not for real ;^)
    – Michael Hoppe
    Nov 22 '18 at 19:28














  • 1




    Your $frac{11}{10}$ can't possibly be the cosine of anything!
    – TonyK
    Nov 22 '18 at 18:05






  • 1




    @TonyK Not for real ;^)
    – Michael Hoppe
    Nov 22 '18 at 19:28








1




1




Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 '18 at 18:05




Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 '18 at 18:05




1




1




@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 '18 at 19:28




@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 '18 at 19:28










5 Answers
5






active

oldest

votes


















3














You cannot separate out the $cos$ function as you have done in step two.



You can remember this identity.



$$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



Using thus result you should get the desired answer.






share|cite|improve this answer



















  • 1




    (+1) ... for discussing the error in the OP
    – Mark Viola
    Nov 22 '18 at 18:05










  • @MarkViola thanks for your upvote.
    – Akash Roy
    Nov 22 '18 at 18:06










  • Thanks once again for your valuable edits @Mark, will bear these in mind.
    – Akash Roy
    Nov 22 '18 at 18:08










  • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
    – LuminousNutria
    Nov 22 '18 at 18:09












  • @Luminous you can accept my answer if you understood.
    – Akash Roy
    Nov 22 '18 at 18:10



















1














The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.






share|cite|improve this answer

















  • 1




    (+1) for mentioning the error in the OP
    – Mark Viola
    Nov 22 '18 at 18:10



















0














$$
cos(a+b)=cos acos b-sin asin b
$$
and hence
$$
cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
$$






share|cite|improve this answer





























    0














    By compound-angle formula,
    begin{eqnarray*}
    cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
    & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
    end{eqnarray*}



    To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
    Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
    and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
    $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
    It follows that
    begin{eqnarray*}
    cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
    & = & frac{3}{10}-frac{2sqrt{3}}{5}
    end{eqnarray*}






    share|cite|improve this answer





























      -1














      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.






      share|cite|improve this answer





















      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
        – Saucy O'Path
        Nov 24 '18 at 10:09












      • ... and $cos 97^circ$ is not $-0.39$.
        – Saucy O'Path
        Nov 24 '18 at 11:17











      Your Answer





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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      You cannot separate out the $cos$ function as you have done in step two.



      You can remember this identity.



      $$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



      Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



      Using thus result you should get the desired answer.






      share|cite|improve this answer



















      • 1




        (+1) ... for discussing the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:05










      • @MarkViola thanks for your upvote.
        – Akash Roy
        Nov 22 '18 at 18:06










      • Thanks once again for your valuable edits @Mark, will bear these in mind.
        – Akash Roy
        Nov 22 '18 at 18:08










      • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
        – LuminousNutria
        Nov 22 '18 at 18:09












      • @Luminous you can accept my answer if you understood.
        – Akash Roy
        Nov 22 '18 at 18:10
















      3














      You cannot separate out the $cos$ function as you have done in step two.



      You can remember this identity.



      $$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



      Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



      Using thus result you should get the desired answer.






      share|cite|improve this answer



















      • 1




        (+1) ... for discussing the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:05










      • @MarkViola thanks for your upvote.
        – Akash Roy
        Nov 22 '18 at 18:06










      • Thanks once again for your valuable edits @Mark, will bear these in mind.
        – Akash Roy
        Nov 22 '18 at 18:08










      • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
        – LuminousNutria
        Nov 22 '18 at 18:09












      • @Luminous you can accept my answer if you understood.
        – Akash Roy
        Nov 22 '18 at 18:10














      3












      3








      3






      You cannot separate out the $cos$ function as you have done in step two.



      You can remember this identity.



      $$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



      Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



      Using thus result you should get the desired answer.






      share|cite|improve this answer














      You cannot separate out the $cos$ function as you have done in step two.



      You can remember this identity.



      $$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



      Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



      Using thus result you should get the desired answer.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 22 '18 at 18:07









      Mark Viola

      130k1274170




      130k1274170










      answered Nov 22 '18 at 18:00









      Akash RoyAkash Roy

      1




      1








      • 1




        (+1) ... for discussing the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:05










      • @MarkViola thanks for your upvote.
        – Akash Roy
        Nov 22 '18 at 18:06










      • Thanks once again for your valuable edits @Mark, will bear these in mind.
        – Akash Roy
        Nov 22 '18 at 18:08










      • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
        – LuminousNutria
        Nov 22 '18 at 18:09












      • @Luminous you can accept my answer if you understood.
        – Akash Roy
        Nov 22 '18 at 18:10














      • 1




        (+1) ... for discussing the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:05










      • @MarkViola thanks for your upvote.
        – Akash Roy
        Nov 22 '18 at 18:06










      • Thanks once again for your valuable edits @Mark, will bear these in mind.
        – Akash Roy
        Nov 22 '18 at 18:08










      • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
        – LuminousNutria
        Nov 22 '18 at 18:09












      • @Luminous you can accept my answer if you understood.
        – Akash Roy
        Nov 22 '18 at 18:10








      1




      1




      (+1) ... for discussing the error in the OP
      – Mark Viola
      Nov 22 '18 at 18:05




      (+1) ... for discussing the error in the OP
      – Mark Viola
      Nov 22 '18 at 18:05












      @MarkViola thanks for your upvote.
      – Akash Roy
      Nov 22 '18 at 18:06




      @MarkViola thanks for your upvote.
      – Akash Roy
      Nov 22 '18 at 18:06












      Thanks once again for your valuable edits @Mark, will bear these in mind.
      – Akash Roy
      Nov 22 '18 at 18:08




      Thanks once again for your valuable edits @Mark, will bear these in mind.
      – Akash Roy
      Nov 22 '18 at 18:08












      Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
      – LuminousNutria
      Nov 22 '18 at 18:09






      Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
      – LuminousNutria
      Nov 22 '18 at 18:09














      @Luminous you can accept my answer if you understood.
      – Akash Roy
      Nov 22 '18 at 18:10




      @Luminous you can accept my answer if you understood.
      – Akash Roy
      Nov 22 '18 at 18:10











      1














      The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.






      share|cite|improve this answer

















      • 1




        (+1) for mentioning the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:10
















      1














      The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.






      share|cite|improve this answer

















      • 1




        (+1) for mentioning the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:10














      1












      1








      1






      The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.






      share|cite|improve this answer












      The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 22 '18 at 17:59









      tarit goswamitarit goswami

      1,7381421




      1,7381421








      • 1




        (+1) for mentioning the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:10














      • 1




        (+1) for mentioning the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:10








      1




      1




      (+1) for mentioning the error in the OP
      – Mark Viola
      Nov 22 '18 at 18:10




      (+1) for mentioning the error in the OP
      – Mark Viola
      Nov 22 '18 at 18:10











      0














      $$
      cos(a+b)=cos acos b-sin asin b
      $$
      and hence
      $$
      cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
      $$






      share|cite|improve this answer


























        0














        $$
        cos(a+b)=cos acos b-sin asin b
        $$
        and hence
        $$
        cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
        $$






        share|cite|improve this answer
























          0












          0








          0






          $$
          cos(a+b)=cos acos b-sin asin b
          $$
          and hence
          $$
          cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
          $$






          share|cite|improve this answer












          $$
          cos(a+b)=cos acos b-sin asin b
          $$
          and hence
          $$
          cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 '18 at 18:03









          Yiorgos S. SmyrlisYiorgos S. Smyrlis

          62.8k1383163




          62.8k1383163























              0














              By compound-angle formula,
              begin{eqnarray*}
              cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
              & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
              end{eqnarray*}



              To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
              Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
              and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
              $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
              It follows that
              begin{eqnarray*}
              cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
              & = & frac{3}{10}-frac{2sqrt{3}}{5}
              end{eqnarray*}






              share|cite|improve this answer


























                0














                By compound-angle formula,
                begin{eqnarray*}
                cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
                & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
                end{eqnarray*}



                To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
                Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
                and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
                $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
                It follows that
                begin{eqnarray*}
                cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
                & = & frac{3}{10}-frac{2sqrt{3}}{5}
                end{eqnarray*}






                share|cite|improve this answer
























                  0












                  0








                  0






                  By compound-angle formula,
                  begin{eqnarray*}
                  cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
                  & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
                  end{eqnarray*}



                  To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
                  Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
                  and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
                  $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
                  It follows that
                  begin{eqnarray*}
                  cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
                  & = & frac{3}{10}-frac{2sqrt{3}}{5}
                  end{eqnarray*}






                  share|cite|improve this answer












                  By compound-angle formula,
                  begin{eqnarray*}
                  cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
                  & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
                  end{eqnarray*}



                  To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
                  Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
                  and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
                  $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
                  It follows that
                  begin{eqnarray*}
                  cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
                  & = & frac{3}{10}-frac{2sqrt{3}}{5}
                  end{eqnarray*}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 '18 at 18:05









                  Danny Pak-Keung ChanDanny Pak-Keung Chan

                  2,25038




                  2,25038























                      -1














                      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.






                      share|cite|improve this answer





















                      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                        – Saucy O'Path
                        Nov 24 '18 at 10:09












                      • ... and $cos 97^circ$ is not $-0.39$.
                        – Saucy O'Path
                        Nov 24 '18 at 11:17
















                      -1














                      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.






                      share|cite|improve this answer





















                      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                        – Saucy O'Path
                        Nov 24 '18 at 10:09












                      • ... and $cos 97^circ$ is not $-0.39$.
                        – Saucy O'Path
                        Nov 24 '18 at 11:17














                      -1












                      -1








                      -1






                      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.






                      share|cite|improve this answer












                      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 24 '18 at 9:59









                      Siddharth DhimanSiddharth Dhiman

                      1




                      1












                      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                        – Saucy O'Path
                        Nov 24 '18 at 10:09












                      • ... and $cos 97^circ$ is not $-0.39$.
                        – Saucy O'Path
                        Nov 24 '18 at 11:17


















                      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                        – Saucy O'Path
                        Nov 24 '18 at 10:09












                      • ... and $cos 97^circ$ is not $-0.39$.
                        – Saucy O'Path
                        Nov 24 '18 at 11:17
















                      The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                      – Saucy O'Path
                      Nov 24 '18 at 10:09






                      The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                      – Saucy O'Path
                      Nov 24 '18 at 10:09














                      ... and $cos 97^circ$ is not $-0.39$.
                      – Saucy O'Path
                      Nov 24 '18 at 11:17




                      ... and $cos 97^circ$ is not $-0.39$.
                      – Saucy O'Path
                      Nov 24 '18 at 11:17


















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